Solve for $x$ : $ 2|x + 1| + 1 = 6|x + 1| + 3 $
Solution: Subtract $ {2|x + 1|} $ from both sides: $ \begin{eqnarray} 2|x + 1| + 1 &=& 6|x + 1| + 3 \\ \\ {- 2|x + 1|} && {- 2|x + 1|} \\ \\ 1 &=& 4|x + 1| + 3 \end{eqnarray} $ Subtract $3$ from both sides: $ \begin{eqnarray} 1 &=& 4|x + 1| + 3 \\ \\ {- 3} && {- 3} \\ \\ -2 &=& 4|x + 1| \end{eqnarray} $ Divide both sides by ${4}$ $ \dfrac{-2} {{4}} = \dfrac{4|x + 1|} {{4}} $ Simplify: $ -\dfrac{1}{2} = |x + 1| $ The absolute value cannot be negative. Therefore, there is no solution.